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An integer x is selected at random from the set {17, 21, 23,
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12 Sep 2019, 04:33
12 Sep 2019, 04:33
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26% (01:17) wrong based on 136 sessions
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An integer x is selected at random from the set {17, 21, 23, 25, 27, 30, 33}.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
The probability that the average (arithmetic mean) of 8, 16, and x is at least 17 |
\(\frac{1}{2}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: An integer x is selected at random from the set {17, 21, 23,
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12 Sep 2019, 10:36
12 Sep 2019, 10:36
2
Carcass wrote:
An integer x is selected at random from the set {17, 21, 23, 25, 27, 30, 33}.
Quantity A |
Quantity B |
The probability that the average (arithmetic mean) of 8, 16, and x is at least 17 |
\(\frac{1}{2}\) |
Let's first determine what x-value will yield an average of EXACTLY 17
So, we can write: \(\frac{8+16+x}{3}=17\)
Multiply both sides by 3 to get: \(8+16+x=51\)
Simplify to get: \(24+x=51\)
Solve: \(x=27\)
So, x=27 yields an average of 17
So P(average is at LEAST 17) = P(selected value is greater than or equal to 27)
In the set {17, 21, 23, 25, 27, 30, 33}, 3 of the 7 values are greater than or equal to 27
So, P(average is at LEAST 17) = P(selected value is greater than or equal to 27) = 3/7
We get:
QUANTITY A: 3/7
QUANTITY B: 1/2
1/2 > 3/7. So, the answer is B
Cheers
Brent
Re: An integer x is selected at random from the set {17, 21, 23,
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01 Feb 2021, 09:24
01 Feb 2021, 09:24
1
Hi,
As we need the average to be 17, So we basically need the sum to be greater than 17*3=51 and currently the sum is 51-8-16=27. Thus any number above or equal to 27 will work. Giving us a probability of 3/7 which is less than 1/2.
IMO B
As we need the average to be 17, So we basically need the sum to be greater than 17*3=51 and currently the sum is 51-8-16=27. Thus any number above or equal to 27 will work. Giving us a probability of 3/7 which is less than 1/2.
IMO B
